LCM and HCF – Quick Revision Notes (JKSSB – Social Forestry Worker – Basic Mathematics)
1. What are LCM and HCF?
| Term | Full Form | Meaning | Symbol |
|---|---|---|---|
| LCM | Least Common Multiple | Smallest positive integer that is exactly divisible by each of the given numbers. | \( \text{LCM}(a,b) \) |
| HCF | Highest Common Factor (also GCD – Greatest Common Divisor) | Largest positive integer that divides each of the given numbers without leaving a remainder. | \( \text{HCF}(a,b) \) or \( \gcd(a,b) \) |
- Key fact: For any two positive integers \(a\) and \(b\)
\[
\boxed{\text{LCM}(a,b)\times\text{HCF}(a,b)=a\times b}
\]
2. Core Concepts & Quick Checks
- Co‑prime numbers – two numbers whose HCF = 1.
- Example: (8, 15), (9, 28).
- For co‑primes: \(\text{LCM}=a\times b\).
- Identical numbers – if \(a=b\):
- \(\text{LCM}=a=b\) and \(\text{HCF}=a=b\). – One of the numbers is 0 (rare in exam but good to know): – \(\text{HCF}(a,0)=|a|\).
- \(\text{LCM}(a,0)=0\) (by convention, because any multiple of 0 is 0).
- Prime numbers – HCF of any two distinct primes = 1; LCM = product of the primes.
- Powers of the same prime – e.g., \(2^3\) and \(2^5\):
- HCF = lower power = \(2^3\).
- LCM = higher power = \(2^5\).
3. Methods to Find HCF
| Method | Steps | When to Use |
|---|---|---|
| Prime Factorisation | 1. Write each number as product of prime powers. 2. Take the lowest power of each common prime. 3. Multiply them. |
Small numbers (< 200) or when factors are obvious. |
| Division (Euclid’s) Algorithm | 1. Divide larger number by smaller → remainder \(r\). 2. Replace divisor with dividend, dividend with remainder. 3. Repeat until remainder = 0. 4. Last non‑zero divisor = HCF. |
Any size; especially useful for large numbers or when factoring is tedious. |
| Short‑cut for Two Numbers | If one number divides the other → HCF = smaller number. | Quick check before doing full work. |
| Using LCM (via the product relation) | HCF = \(\dfrac{a\times b}{\text{LCM}(a,b)}\) – compute LCM first if easier. | When LCM is obvious (e.g., numbers are co‑prime). |
Mnemonic for Euclid’s algorithm:
“Divide, Swap, Repeat till Zero – the last divisor is the HCF.”
4. Methods to Find LCM
| Method | Steps | When to Use |
|---|---|---|
| Prime Factorisation | 1. Prime‑factorise each number. 2. For each prime, take the highest power appearing in any number. 3. Multiply all selected powers. |
Numbers with small prime bases; good for verification. |
| Division Method (Ladder) | 1. Write numbers in a row. 2. Divide by any prime that exactly divides at least two numbers. 3. Write quotients (and undivided numbers) below. 4. Repeat until no prime divides two or more numbers. 5. LCM = product of all divisors used × product of numbers in the last row. |
Faster than pure factorisation for 3‑4 numbers; visual. |
| Using HCF | LCM = \(\dfrac{a\times b}{\text{HCF}(a,b)}\) – compute HCF first if easier. | When HCF is obvious (e.g., one number divides the other). |
| Listing Multiples (only for very small numbers) | List first few multiples of each number; the first common one is LCM. | Quick sanity check; not recommended for exam timing. |
Mnemonic for Ladder method:
“Prime‑divide, bring down, keep going – multiply the divisors and the leftovers.”
5. Important Properties & Formulas
| Property | Statement | Example |
|---|---|---|
| Commutative | \(\text{LCM}(a,b)=\text{LCM}(b,a)\) ; \(\text{HCF}(a,b)=\text{HCF}(b,a)\) | LCM(4,6)=LCM(6,4)=12 |
| Associative | \(\text{LCM}(a,\text{LCM}(b,c))=\text{LCM}(\text{LCM}(a,b),c)\) (same for HCF) | LCM(2,LCM(3,4))=LCM(LCM(2,3),4)=12 |
| Distributive over multiplication | \(\text{LCM}(ka, kb)=k\cdot\text{LCM}(a,b)\) ; \(\text{HCF}(ka, kb)=k\cdot\text{HCF}(a,b)\) | LCM(6,10)=2·LCM(3,5)=2·15=30 |
| Relation with product | \(\text{LCM}(a,b)\times\text{HCF}(a,b)=a\times b\) | For 12 and 18: LCM=36, HCF=6 → 36·6=216=12·18 |
| LCM of fractions | \(\text{LCM}\left(\frac{a}{b},\frac{c}{d}\right)=\frac{\text{LCM}(a,c)}{\text{HCF}(b,d)}\) | LCM\(\left(\frac{2}{3},\frac{3}{4}\right)=\frac{\text{LCM}(2,3)}{\text{HCF}(3,4)}=\frac{6}{1}=6\) |
| HCF of fractions | \(\text{HCF}\left(\frac{a}{b},\frac{c}{d}\right)=\frac{\text{HCF}(a,c)}{\text{LCM}(b,d)}\) | HCF\(\left(\frac{2}{3},\frac{3}{4}\right)=\frac{\text{HCF}(2,3)}{\text{LCM}(3,4)}=\frac{1}{12}\) |
| LCM of more than two numbers | Apply pairwise or use prime‑factorisation: take max power of each prime across all numbers. | LCM(4,6,9)=2²·3²=36 |
| HCF of more than two numbers | Apply pairwise or use prime‑factorisation: take min power of each prime common to all numbers. | HCF(12,18,24)=2·3=6 |
6. Step‑by‑Step Worked Examples #### Example 1 – HCF by Euclid’s algorithm
Find HCF(252, 105).
- \(252 ÷ 105 = 2\) remainder \(42\). 2. \(105 ÷ 42 = 2\) remainder \(21\).
- \(42 ÷ 21 = 2\) remainder \(0\).
Last non‑zero divisor = 21 → HCF = 21.
(Check: \(252=21×12\), \(105=21×5\)).
Example 2 – LCM by Ladder method
Find LCM(24, 36, 60).
2 | 24 36 60
2 | 12 18 30 2 | 6 9 15
3 | 2 3 5
5 | 2 1 1
Divisors used: \(2,2,2,3,5\) → product = \(2×2×2×3×5 = 120\).
Remaining numbers in last row: \(2,1,1\) → product = 2.
LCM = \(120 × 2 = 240\).
(Verification: 240 is divisible by 24, 36, 60.)
Example 3 – Using the product relation Given HCF(48, 180) = 12, find LCM.
\[
\text{LCM}= \frac{48×180}{12}= \frac{8640}{12}=720.
\]
Check: LCM×HCF = 720×12 = 8640 = 48×180 ✔ #### Example 4 – Fractions
Find LCM and HCF of \(\frac{5}{6}\) and \(\frac{7}{9}\).
- LCM: \(\frac{\text{LCM}(5,7)}{\text{HCF}(6,9)} = \frac{35}{3}= 11\frac{2}{3}\).
- HCF: \(\frac{\text{HCF}(5,7)}{\text{LCM}(6,9)} = \frac{1}{18}\).
7. Quick‑Reference Tables
Table A – LCM & HCF of Small Number Pairs (1‑12)
| a | b | HCF | LCM |
|---|---|---|---|
| 1 | any | 1 | b |
| 2 | 4 | 2 | 4 |
| 2 | 6 | 2 | 6 |
| 2 | 9 | 1 | 18 |
| 3 | 5 | 1 | 15 |
| 3 | 9 | 3 | 9 |
| 4 | 6 | 2 | 12 |
| 4 | 10 | 2 | 20 |
| 5 | 7 | 1 | 35 |
| 6 | 8 | 2 | 24 |
| 6 | 9 | 3 | 18 |
| 7 | 14 | 7 | 14 |
| 8 | 12 | 4 | 24 |
| 9 | 12 | 3 | 36 |
| 10 | 15 | 5 | 30 |
| 11 | 13 | 1 | 143 |
| 12 | 18 | 6 | 36 |
Use this table to spot patterns: when one number divides the other, HCF = smaller, LCM = larger.
Table B – Prime‑Power Decomposition (up to 20)
| Number | Prime Factorisation |
|---|---|
| 2 | \(2^1\) |
| 3 | \(3^1\) |
| 4 | \(2^2\) |
| 5 | \(5^1\) |
| 6 | \(2^1·3^1\) |
| 7 | \(7^1\) |
| 8 | \(2^3\) |
| 9 | \(3^2\) |
| 10 | \(2^1·5^1\) |
| 11 | \(11^1\) |
| 12 | \(2^2·3^1\) |
| 13 | \(13^1\) |
| 14 | \(2^1·7^1\) |
| 15 | \(3^1·5^1\) |
| 16 | \(2^4\) |
| 17 | \(17^1\) |
| 18 | \(2^1·3^2\) |
| 19 | \(19^1\) |
| 20 | \(2^2·5^1\) |
Having this table memorised speeds up both HCF (take min powers) and LCM (take max powers).
8. Mnemonics & Memory Tricks
| Concept | Mnemonic | How it helps |
|---|---|---|
| HCF vs LCM | “HCF = Highest Common Factor → think ‘small’ (the biggest that still fits into both). LCM = Lowest Common Multiple → think ‘large’ (the smallest that both can reach).” |
Visual contrast: HCF ≤ numbers ≤ LCM. |
| Co‑prime clue | “If they share no prime, HCF = 1, LCM = product.” | Quick test for co‑primes. |
| Euclid’s steps | “Divide, Swap, Repeat till Zero – last divisor is HCF.” | Recalls the algorithm in order. |
| Ladder (division) method | “Prime‑divide, bring down, keep going – multiply the divisors and the leftovers.” | Reminds you to multiply all used primes and the final row. |
| Product relation | “LCM × HCF = Product – think of a rectangle: area = length × breadth.” | Links LCM & HCF to the original numbers. |
| Fractions LCM/HCF | “LCM of fractions → LCM of numerators over HCF of denominators. HCF of fractions → HCF of numerators over LCM of denominators.” |
Flips the operation for numerators vs denominators. |
| Powers of same prime | “HCF takes the lowest exponent, LCM takes the highest exponent.” | Directly applies to \(p^a, p^b\). |
9. Common Pitfalls & How to Avoid Them
| Mistake | Why it Happens | Correct Approach |
|---|---|---|
| Confusing HCF with LCM (e.g., giving LCM when asked HCF) | Similar names; both involve “common”. | Recall: HCF ≤ each number; LCM ≥ each number. If answer is smaller than the given numbers → HCF; if larger → LCM. |
| Forgetting to reduce fraction before LCM/HCF | Working with raw numerators/denominators. | Always reduce fractions to lowest terms first; then apply the LCM/HCF formulas for fractions. |
| Using product relation when numbers are not integers | Applying \(LCM×HCF = a×b\) to fractions directly. | Convert fractions to the form \(\frac{p}{q}\) and use the fraction‑specific formulas, or convert to a common denominator first. |
| Missing a prime factor in ladder method | Stopping division too early when a prime divides only one number. | Continue dividing by any prime that divides at least two numbers; if none, bring down the untouched numbers as they are. |
| Mis‑applying the associative property | Trying to group numbers incorrectly for more than two numbers. | For three or more numbers, you can iteratively apply binary LCM/HCF: \(\text{LCM}(a,b,c)=\text{LCM}(\text{LCM}(a,b),c)\). Same for HCF. |
| Assuming HCF of two even numbers is always 2 | Overgeneralising. | HCF depends on common prime powers; e.g., HCF(12,18)=6, not 2. Always factorise or use Euclid. |
| Using listing multiples for large numbers | Time‑consuming and error‑prone. | Reserve listing only for quick checks on numbers ≤ 12; otherwise use prime factorisation or ladder method. |
10. Practice Questions (with Answers)
Solve quickly – aim for < 2 minutes each.
- Find HCF and LCM of 84 and 126.
- HCF(45, 75) = ? Hence find LCM(45,75).
- Three bells toll at intervals of 6 s, 8 s and 12 s. After how many seconds will they toll together again?
- The product of two numbers is 2160 and their HCF is 12. Find their LCM.
- Find LCM of \(\frac{3}{4}\) and \(\frac{5}{6}\).
- Find HCF of \(\frac{2}{3}\), \(\frac{4}{9}\) and \(\frac{5}{12}\).
- Using Euclid’s algorithm, compute HCF(275, 130).
- Using the ladder method, find LCM of 16, 24 and 36.
- If LCM of two numbers is 180 and their product is 5400, what is their HCF?
- Two numbers are in the ratio 5:7 and their LCM is 210. Find the numbers.
Answers
- HCF = 42, LCM = 252
- HCF = 15 → LCM = \(\frac{45×75}{15}=225\)
- LCM(6,8,12)=24 s
- LCM = \(\frac{2160}{12}=180\)
- LCM = \(\frac{\text{LCM}(3,5)}{\text{HCF}(4,6)}=\frac{15}{2}=7\frac{1}{2}\)
- HCF = \(\frac{\text{HCF}(2,4,5)}{\text{LCM}(3,9,12)}=\frac{1}{36}\)
- Steps: 275÷130=2 r15; 130÷15=8 r10; 15÷10=1 r5; 10÷5=2 r0 → HCF=5
- Ladder → divisors: 2,2,2,3 → product=48; last row: 2,3,3 → product=18 → LCM=48×18=864
- HCF = \(\frac{Product}{LCM}= \frac{5400}{180}=30\)
- Let numbers be 5x, 7x. LCM = 35x (since 5 and 7 are co‑prime). 35x=210 → x=6 → numbers = 30 and 42.
11. Last‑Minute Revision Checklist
- [ ] Recall the definition of HCF and LCM in one sentence.
- [ ] Write down the product relation \(\text{LCM}×\text{HCF}=a×b\).
- [ ] Know when numbers are co‑prime → HCF=1, LCM=product.
- [ ] Remember Euclid’s steps: Divide → Swap → Repeat → Zero.
- [ ] Remember ladder method: Divide by any prime that hits ≥2 numbers, bring down, repeat.
- [ ] For fractions: LCM of numerators / HCF of denominators; HCF of numerators / LCM of denominators.
- [ ] For powers of the same prime: HCF = lower exponent, LCM = higher exponent. – [ ] Use the product relation to find the missing value when HCF or LCM is known.
- [ ] Avoid common pitfalls (confusing HCF/LCM, forgetting to reduce fractions, prematurely stopping ladder).
- [ ] Quick‑scan the small‑pair table (1‑12) for instant recognition.
- [ ] Practice at least two problems from each method before the exam.
You’re now equipped with a full‑spectrum, exam‑ready refresher on LCM and HCF.
Go through the definitions, work through a couple of examples using each method, and revise the mnemonics. With these tools, tackling any LCM/HCF question in the JKSSB Social Forestry Worker paper will be swift and accurate. Good luck!
